DESPERATE NEED OF HELP ON THIS Physics PROBLEM

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The discussion revolves around a physics problem involving a police car and a speeder, where the police car accelerates after a delay. The key steps to solve the problem include converting speeds from km/h to m/s, calculating the initial distance the speeder gains during the police car's delay, and setting up equations for the distances traveled by both vehicles. Participants emphasize the importance of guiding the original poster (OP) to understand the problem rather than providing direct answers, adhering to forum guidelines. Ultimately, the solution involves using the quadratic formula to find the time it takes for the police car to overtake the speeder. The community aims to foster learning through hints and suggestions.
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DESPERATE NEED OF HELP ON THIS Physics PROBLEM!

Okay for those who are really smart in Physics, please help! I've been staring at this problem for hours and I just can't figure it out. :rolleyes: I know that you need to find the final velocity and then time, but it just seems like there's not enough information. Well here's the problem:

An unmarked police car, traveling a constant 90 km/h, is passed by a speeder traveling 140 km/h. Precisely 1.00s after the speeder passes, the policeman steps on the accelerator. If the police car's acceleration is 2.00 m/s2, how much time elapses after the police car is passed until it over takes the speeder (assumed moving at constant speed)?

I don't need the answer or anything, I just need to know where to start off 'cause I'm totally stuck. If you figure it out please email, message, or even IM me on MSN at pesstimisticgurl23@yahoo.com.Thank you so much!
 
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Write an equation for the distance between the two cars. Start with the one second delay -- how far does that put the speeder ahead of the cop at the instant that the cop car begins its acceleration? And in that second phase (cop car accelerating), what is the equation for the relative distance between the two cars?
 
First, change km/h to m/s for convenience.
Then set up the equation x = Vit + (1/2)at^2 for each object:

Police car:
x + 13.889 = 25t + t^2 ------> x = 25t + t^2 - 13.889

Speeder:
x = 38.889t


Now, make x of police car and x of speeder equal each other and you get:
25t + t^2 - 13.889 = 38.889t ------> t^2 - 13.889t - 13.889


Solve for time by using the quadratic formula. This is the instance that the two objects meet each other.

Once you do the calculation, you get 14.8 second as an answer.
 
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ysk1, We are not supposed to give out the answers to problems here in the Homework Forums. That is part of the guidelines that you agreed to in order to get into these HW forums. Our job is to give suggestions, hints and other tips to help the original poster (OP) to figure out the problem on their own. There is very little value to the OP to just post the answer. They need to figure out most of it on their own in order to learn. Please do not post the answer in any other threads in the Homework Forums.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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